Practice Quiz with Solutions: Chapter 5 (50 min)

Disclaimer: This quiz is representative of the level of
difficulty you should expect, but it doesn’t include every single
topic from the week’s work. The real quiz may include some other
topics and may skip some that are in this practice quiz. (The real
quiz also may not word questions in the same way as the practice
quiz. You should focus on the concepts, not a particular form of
words.)

Common mistake:
Sometimes students write the sample space correctly but
miss one of the combinations of 2 heads. I wish I could
offer some “magic bullet” for counting correctly, but the only advice
I have is just to be really careful.

3(points: 3) P(A), the probability of event A, is 0.7. A and B are
complementary events.
(a) Find P(not A).
(b) Find P(B).
(c) Find P(A and B).
If any of the above cannot be determined from the information
given, say so.

Solution:
(a) P(not A) = 1−P(A) = 1−0.7
→ P(not A) = 0.3 (b) That A and B are complementary means that one or the other
must happen, but not both. Therefore P(B) = P(not A)
→ P(B) = 0.3 (c) Since the events are complementary, they can’t both happen:
P(A and B) = 0

Common mistake:
Many students get (c) wrong, giving an answer of 1. If events
are complementary, they can’t both happen at the same time. That means
P(A and B) must be 0, the probability of something
impossible.

Maybe those students were thinking of P(A or B). It’s true that
if A and B are complementary then P(A or B) =
P(A) + P(B) = 1. But part (c) was about probability
and, not probability or.

4(points: 2) You draw a card at random from a standard 52-card deck.
What’s the probability that it’s an ace or a spade?

Solution:
“Ace” and “spade” are non-disjoint events,
since a given card can be both an ace and a spade.
Therefore, you don’t have a formula for this problem.
But you can go back to the sample space and
compute the probability by counting.

There are thirteen spades in the deck, and three aces in addition to
the ace of spades, which you already counted as one of the spades.
13+3 = 16, out of 52 cards. P(Ace or Spade) =
16/52

5(points: 2) In Monopoly, if you roll doubles you get an extra roll, but if
you roll doubles three times in a row you go to jail.
(a) On any given turn, what’s the probability you’ll
roll doubles? (The
picture in
the handout may help.)
(b) On any given turn, what’s the probability you’ll
go to jail for rolling doubles three times in a row?

Solution:
(a) There are 6 ways to roll doubles, and the sample space is
36 equally likely events.

P(doubles on 1 roll) = 6/36 = 1/6

(b) Successive rolls are independent events, so you use the
simple multiplication rule:

6(points: 2) At a job fair, 200 students comprised 62 nursing majors, 45
business majors, 12 hospitality, 8 construction, 10 science, and 63 liberal arts.
If you randomly select a student, what’s the probability that he
or she is majoring in nursing or liberal arts?

Solution:62+45+12+8+10+63 = 200, so you know
that nobody has two majors and therefore all the majors are disjoint
events. That lets you use the simple addition rule:

P(N or LA) = P(N) + P(LA)

To find those probabilities, you might find it easier to think of them
as proportions of the group instead of probabilities for one student.

P(N or LA) = P(N) + P(LA)

P(N or LA)= 62/200 + 63/200 = 125/200 or 5/8 or 0.625

Registrations (thousands)

Male

Female

Total

Democrat

21

28

49

Republican

24

21

45

Green

2

3

5

other

0

1

1

Total

47

53

100

Voter registrations in a fictional New York county are
summarized in the table.
Use the table to answer these questions about a
randomly selected registered voter:

7(points: 2) P(Republican or Female) =

Solution:
The events “Republican” and “Female” are not disjoint, so you
add 53 thousand females plus the 24 thousand male Republicans, to get
77 thousand who are Republican or female. P(Republican or
female) = 77/100 or 77%.

8(points: 2) P(Republican and Female) =

Solution:
The denominator again is 100, the total registrations. The
numerator is 21, the number of female Republicans.

P(Republican and Female) = 21/100 = 0.21 or 21%

Common mistake:
You don’t need for any kind of formula here. The
“and”condition is already covered by the fact that the cell
containing 21 is the intersection of the Republican row and the Female
column.

Common mistake:
Some students try to multiply P(Republican) times P(Female).
This is wrong, because the simple multiplication rule applies only
when the events are independent. Here the events
“Republican” and “Female” are not independent, so
the short form of the rule for “and” does not apply.

9(points: 2) What’s the probability that a randomly selected female
is a Republican?

Solution:
There are 53 thousand females, and 21 thousand of them are
Republicans, so the probability is
21/53 or about 39.62%

10(points: 4) Tom Turkey invested in two stocks, A and W. There is a
0.90 probability that company A will go bankrupt, and a 0.80
probability that company W will go bankrupt. Assuming the two
companies have no connection, find the probabilities that
(a) both will go bankrupt; (b) one of
them, but not both, will go bankrupt; (c) neither will go
bankrupt.

Solution:
You’re being asked about all three possibilities: two fail,
one fails, none fail. Therefore the three probabilities must add up
to 1, and you need to compute only two of them.
It’s also important to note that the companies are
independent: whether one fails has nothing to do with whether the
other fails.

(a) Since the companies are independent, you can use the simple
multiplication rule:

P(A bankrupt and W bankrupt) = P(A bankrupt) × P(W bankrupt)

P(A bankrupt and W bankrupt) = .9 × .8 = 0.72

At this point you could compute (b), but it’s
hard: you need the probability that A fails and W is okay, plus the
probability that A is okay and W fails. (c) looks easier, so do that
first.

(c) “Neither bankrupt” means both are okay. Again,
the events are independent so you can use the simple multiplication
rule.

P(neither bankrupt) = P(A okay and W okay)

P(A okay) = 1−.9 = 0.1; P(W okay) = 1−.8 =
0.2

P(neither bankrupt) = .1 × .2 = 0.02

(b) is now a piece of cake.

P(only one bankrupt) = 1 − P(both bankrupt) − P(none bankrupt)

P(only one bankrupt) = 1 − .72 − .02 =
0.26

Solution:
If you have time, it’s always good to check your work
using a different method. You can work out (b) the long way.
You have only independent events (whether A is okay or fails, whether W
is okay or fails) and disjoint events (A fails and W okay, A okay and
W fails). The “okay” probabilities were computed in part
(c).

Common mistake:
A very common mistake is to solve only half the problem when
working this out the long way. When
you have probability of one not the other, you have to consider both
A-and-not-W and W-and-not-A.

Remark:
If you computed all three probabilities the long way, pause a
moment to check your work by adding them to make sure you get 1.
Whenever possible,
check your work with a second type of computation.

11(points: 2) A poll found that 45% of baseball fans had attended a game in
person within the past year. Of five randomly selected baseball fans,
find the probability that at least one had not attended a
game within the past year.

Solution:
In “at least” and “no more than”
probability problems, the complement is often your friend. The
complement of “at least one had not attended” is “all
had attended”. If the fans are randomly selected, their opinions
are independent and you can use the simple multiplication rule.

P(all 5 attended) = 0.45^5 = 0.0185

P(at least 1 had not attended) = 1 − 0.0185 =
0.9815

12(points: 2) A second poll found that 86% of baseball fans favored drug testing for
baseball players. What’s the
probability that a given baseball fan had attended a game within the
past year and favored drug testing?

Solution:
This is a probability “and”, so you need independent
events. But you don’t have any basis to think that fans who
attend games in person have the same opinions as fans who do not. In
other words, you don’t know that “attended a game within
the past year” and “favors drug testing” are
independent events. Answer:
events may not be independent; can’t solve

13(points: 3) In 2003 a federal government survey estimated that 58.2% of US
households had both a cell phone and a landline, 2.8% had only cell
service, and 1.6% had no phone service at all.
(a) Construct a probability model for type of phone service to US
households.
(b) Polling agencies try (in theory) not to call cell phones,
because consumers object to paying for the calls. What proportion of
US households could be reached by a landline in 2003?

Service type

Prob.

Landline only

37.4%

Landline and cell

58.2%

Cell only

2.8%

No phone

1.6%

Total

100.0%

Solution:
(a) In a probability model, the probabilities must add to 1
(= 100%). The given probabilities add to 62.6%. What is the
missing 37.4%? They’ve accounted for cell and landline, cell
only, and nothing; the remaining possibility is landline only. The
model is shown at right.

(b)

P(Landline) = P(Landline only) + P(Landline and cell)

P(Landline) = 37.4% + 58.2% = 95.6%

Remark:
“Landline” and “cell” are not disjoint events,
because a given household could have both.
But “landline only” and “landline and cell” are
disjoint, because a given house can’t both have a cell phone
with landline and have no cell phone with landline.

Remove the problem about VCR and Tivo, because it nearly
duplicated the one about the stocks.

21 Jun 2012: Rule out double majors
here and
here; thanks to Lotte Hammond for pointing
out the ambiguity. Also, simplify the explanation of the
recording devices (and the entire problem would be removed four months
later).